sebanyak 1,6gram NaOH dilarutkan dalam 500 gram air. Titik didih larutan NaOH adalah.... (Kb air=0,25 derajat C/m, Ar:Na=23,O=16,H=1

massa NaOH = 1,6 gram
massa air = 500 gram
kb air = 0,52°C/m (Kb air bukan 0,25 tapi 0,52)
(Ar Na = 23, O = 16, H = 1)

Titik didih larutan / Tb ... ?

*Mr NaOH
Mr = Ar Na + Ar O + Ar H
Mr = 23 + 16 + 1
Mr = 40 gram/mol

*molalitas larutan
m = massa NaOH/Mr NaOH . 1000/massa air
m = 1,6/40 . 1000/500
m = 0,08 molal

*∆Tb = Kb . m
∆Tb = 0,52 . 0,08
∆Tb = 0,0416

*Titik didih larutan NaOH
∆Tb = Tb - Tb°
Tb = ∆Tb + Tb°
Tb = 0,0416 + 100
Tb = 100,0416°C